Problem: Balance the following chemical equation: $ $ $\text{Al} +$ $\text{HCl} \rightarrow$ $\text{AlCl}_3 +$ $\text{H}_2$
Answer: There is $1 \text{ H}$ and $1 \text{ Cl}$ on the left and $3 \text{ Cl}$ and $2 \text{ H}$ on the right. The lowest common denominator for the right is $6$ , so multiply $\text{AlCl}_3$ by ${2}$ and $\text{H}_2$ by ${3}$ $ \text{Al} + \text{HCl} \rightarrow {2}\text{AlCl}_3 + {3}\text{H}_2 $ That gives us $6 \text{ H}$ and $6 \text{ Cl}$ on the right, so multiply $\text{HCl}$ by ${6}$ $ \text{Al} + {6}\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 $ That gives us $2 \text{ Al}$ on the right and only $1$ on the left, so multiply $\text{Al}$ by ${2}$ $ {2}\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 $ The balanced equation is: $ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 $